Ito Integral surjective?

Question

Let $\Phi\in\mathcal{L}\left(M\right)$ if and only if $\Phi$ is a real predictable process and for every $\left\Vert \Phi\right\Vert_{2,t,M}:=\mathbb{E}\left[\int_{0}^{t}\Phi_{s}^2 d\langle M\rangle_{s}\right]^{\frac{1}{2}}<\infty$. And let $\mathcal{M}_{2}$ be the space of square integrable right continuous martingales. It is well known by Ito Isometry that the mapping $$I_{M}:(\mathcal{L}\left(M\right),\left\Vert .\right\Vert_{2})\rightarrow\left(\mathcal{M}_{2},\left|\left\Vert \right|\right\Vert\right)$$ is injective, where $I_{M}\left(\Phi\right)$ is the Ito Integral of $\Phi$.I would like to know if it is also surjective.

Answer 1

No, in general the mapping is not surjective. Consider for example a Brownian motion $(B_t)_{t \geq 0}$. The Itô integral $I_B(\Phi)$ is a continuous martingale and this means that for any (purely) discontinuous martingale $M$ there does not exist $\Phi \in \mathcal{L}(B)$ such that $I_B(\Phi)=M$.

Let me remark that a vast class of (càdlàg) martingales can be represented as a sum of stochastic integrals with respect to a Brownian motion and a Poisson random measure, respectively. This result is due to Ikeda-Watanabe.

Let $(M_t)_{t \geq 0} \in \mathcal{M}_2$ a martingale with respect to a filtration $(\mathcal{F}_t)_{t \geq 0}$ generated by a Lévy process. Then there exists predictable processes $f,g$ as well as a Brownian motion $(B_t)_{t \geq 0}$ and a Poisson random measure $N$ such that $$M_t - M_0 = \int_0^t f(s) \, dB_s + \int_0^t g(s) \, d\tilde{N}_s$$ where $\tilde{N}$ denotes the compensated Poisson random measure.